Problem: Factor completely. $100-9x^2=$
Both $100$ and $9x^2$ are perfect squares, since $100=({10})^2$ and $9x^2=({3x})^2$. $100-9x^2 = ({10})^2-({3x})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={10}$ and ${b}={3x}$ : $({10})^2 - ({3x})^2 =({10}+{3x})({10}-{3x})$ In conclusion, $100-9x^2=(10+3x)(10-3x)$ Remember that you can always check your factorization by expanding it.